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(B) For the quadratic equations,we have:$\alpha_1 + \alpha_2 = -\frac{b}{a}, \alpha_1\alpha_2 = \frac{c}{a}$$\beta_1 + \beta_2 = -\frac{q}{p}, \beta_1

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$b^2pr = q^2ac$

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(B) For the quadratic equations,we have:$\alpha_1 + \alpha_2 = -\frac{b}{a}, \alpha_1\alpha_2 = \frac{c}{a}$$\beta_1 + \beta_2 = -\frac{q}{p}, \beta_1\beta_2 = \frac{r}{p}$The system of linear equations $\alpha_1y + \alpha_2z = 0$ and $\beta_1y + \beta_2z = 0$ has a non-zero solution if the determinant of the coefficient matrix is zero:$\begin{vmatrix} \alpha_1 & \alpha_2 \ \beta_1 & \beta_2 \end{vmatrix} = 0$$\Rightarrow \alpha_1\beta_2 - \alpha_2\beta_1 = 0 \Rightarrow \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} = k$ (say)Using the property of ratios,$\frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} = \frac{\alpha_1 - \alpha_2}{\beta_1 - \beta_2} = k$Also,$\frac{\alpha_1\alpha_2}{\beta_1\beta_2} = k^2$Therefore,$\left( \frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} ight)^2 = \frac{\alpha_1\alpha_2}{\beta_1\beta_2}$Substituting the values:$\left( \frac{-b/a}{-q/p} ight)^2 = \frac{c/a}{r/p}$$\frac{b^2 p^2}{a^2 q^2} = \frac{cp}{ar}$$b^2 p^2 ar = a^2 q^2 cp$$b^2 pr = q^2 ac$

If the roots of the equations $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$ are $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ respectively,and the system of linear equations $\alpha_1y + \alpha_2z = 0$ and $\beta_1y + \beta_2z = 0$ has a non-zero solution,then which of the following is true?

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